Class 12 Physics Atom NCERT Exercise Questions and Answers: Free PDF

12.10 Radius of the Earth's orbit around the Sun, r = 1.5 $*{10}^{11}$ m

Orbital speed of Earth, v = 3 $*{10}^4$ m/s

Mass of the Earth, m = 6 $*{10}^{24}$ kg

According to Bohr's model, angular momentum is given as:

$mvr$ = $\frac{nh}{2?}$ , where

h = Planck's constant = 6.626 $*{10}^{-34}$ Js

n = Quantum number

Hence, n= $\frac{2?mvr}{h}$ = $\frac{2*?*6\mathrm{}*{10}^{24}*3\mathrm{}*{10}^4*1.5*{10}^{11}}{6.626\mathrm{}*{10}^{-34}}$ = 2.56 $*{10}^{74}$

Hence, the quantum number that characterizes earth's revolution is 2.6 $*{10}^{74}$

12.9 It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is – 13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e. -1.1 eV

Orbital energy is related to orbit level(n) as:

E = $\frac{-13.6}{n^2}$ eV

For n = 3, E = $\frac{-13.6}{9}$ eV = - 1.5 eV

This energy is approximately equal to the energy of the gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During the de-excitation, the electron can jump from n=3 to n=1 directly, which forms a line of the Lyman series of hydrogen spectrum.

We have the relation for wave number for Lyman series as:

$\frac{1}{?}$ = $R_y\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$ where

$R_y$ = Rydberg constant = 1.097 $\times {10}^7{m}^{-1}$

$?=$ Wavelength of radiation emitted by the transition of electron

For n = 3, we get

$\frac{1}{?}$ = 1.097 $\times {10}^7\times \left(\frac{1}{1^2}-\frac{1}{3^2}\right)$

$?=1.215\times {10}^{-7}m$ = 102.6 nm

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

$\frac{1}{?}$ = 1.097 $\times {10}^7\times \left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

$?=1.215\times {10}^{-7}m$ = 121.5 nm

If the transition takes place from n = 3 to n = 2, then the wavelength is given as:

$\frac{1}{?}$ = 1.097 $\times {10}^7\times \left(\frac{1}{2^2}-\frac{1}{3^2}\right)$

$?=6.563\times {10}^{-7}m$ = 656.3 nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e. 102.6 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e. 656.3 nm is emitted.

12.8 The radius of the innermost orbit of a hydrogen atom, $r_1$ = 5.3 $*{10}^{-11}$ m

Let $r_2$ be the radius of the orbit at n = 2. The relation between the radius of the orbit is

$r_2$ = $n^2*r_1$ =4 $*5.3*{10}^{-11}\mathrm{}\mathrm{m}$ =21.2 $*{10}^{-11}\mathrm{}\mathrm{m}$

Let $r_3$ be the radius of the orbit at n = 3. The relation between the radius of the orbit is

$r_3$ = $n^2*r_1$ =9 $*5.3*{10}^{-11}\mathrm{}\mathrm{m}$ =47.7 $*{10}^{-11}\mathrm{}\mathrm{m}$

12.1 The size of the atom in Thomson's model is no different from the atomic size in Rutherford's model.

In the ground state of Thomson's model, electrons are in stable equilibrium. While in Rutherford's model, electrons always experience a net force.

A classical atom based on Rutherford's model, is doomed to collapse.

An atom has a nearly continuous mass distribution in a Thomson's model, but has a highly non-uniform mass distribution in Rutherford's model.

The positively charged part of the atom possesses most of the mass in both the models.

12.5 Ground state energy of hydrogen atom, E = -13.6 eV

Kinetic energy is equal to the negative of the total energy (ground state energy) = 13.6 eV

Potential energy is equal to the negative two times of kinetic energy = -13.6 $*2=-27.2eV$

12.4 Separation of two energy level of atom, E = 2.3 eV = 2.3 $*1.6*{10}^{-19}$ J = 3.68 $*{10}^{-19}$

Let $?$ be the frequency of radiation emitted when the atom transits from upper level to lower level.

We have the relation for energy as $E=h?$  , where

h = Planck's constant = 6.626 $*{10}^{-34}$ Js

Then $?=\frac{E}{h}$ = $\frac{3.68*{10}^{-19}}{6.626*{10}^{-34}}$ Hz = 5.55 $*{10}^{14}$ Hz

Hence the frequency is 5.55 $*{10}^{14}$ Hz

12.7 Let $v_1$ be the speed of the electron of the hydrogen atom in the ground state level, $n_1=1.$ For charge e of an electron, $v_1$ is given by the relation

$v_1=\frac{e^2}{n_{1}4?{?}_0\left(\frac{h}{2?}\right)}$ = $\frac{e^2}{2n_1{?}_{0}h}$

$where$ , e = 1.6 $\times {10}^{-19}$ C 

${?}_0=$ Permittivity of free space = 8.85 $\times {10}^{-12}$ $N^{-1}{C}^2{m}^{-2}$

$h$ = Planck’s constant = 6.626 Js $\times {10}^{-34}$

$Hence$ $v_1=\frac{e^2}{2n_1{?}_{0}h}$ $\frac{{(1.6\times {10}^{-19})}^2}{2\times 1\times 8.85\times {10}^{-12}\times 6.626\times {10}^{-34}}$ = 2.182 $\times {10}^6$ m/s

For level 2, $n_2=2$ ,

$v_2=\frac{e^2}{2n_2{?}_{0}h}$ = $\frac{{(1.6\times {10}^{-19})}^2}{2\times 2\times 8.85\times {10}^{-12}\times 6.626\times {10}^{-34}}$ = 1.091 $\times {10}^6$ m/s

For level 3, $n_3=3$ ,

$v_3=\frac{e^2}{2n_3{?}_{0}h}$ = $\frac{{(1.6\times {10}^{-19})}^2}{2\times 3\times 8.85\times {10}^{-12}\times 6.626\times {10}^{-34}}$ = 0.727 $\times {10}^6$ m/s

Let be the orbital period of the electron when it is in level . Orbital period is given by the expression

$T_1$ = $\frac{2?r_1}{v_1}$ where $r_1$ = radius of the orbit = $\frac{n_1^2{h}^2{?}_0}{?m{e}^2}$ , where

$m=$ mass of an electron = 9.1 $\times {10}^{-31}$ kg

e = 1.6 $\times {10}^{-19}$ C

${?}_0=$ Permittivity of free space = 8.85 $\times {10}^{-12}$ $N^{-1}{C}^2{m}^{-2}$

$h$ = Planck’s constant = 6.626 $\times {10}^{-34}$ Js

$T_1$ = $\frac{2?}{v_1}\times \frac{n_1^2{h}^2{?}_0}{?m{e}^2}$ = $\frac{2\times n_1^2\times h^2\times {?}_0}{v_1\times m\times e^2}$ = $\frac{2\times 1\times {(6.626\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{2.182\times {10}^6\times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$ = $\frac{7.77\times {10}^{-78}}{5.083\times {10}^{-62}}$

=1.53 $\times {10}^{-16}s$

For level $n_2$ =2, $T_2$ = $\frac{2\times n_2^2\times h^2\times {?}_0}{v_2\times m\times e^2}$ = $\frac{2\times 4\times {(6.626\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{1.091\times {10}^6\times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$ = $\frac{3.108\times {10}^{-77}}{2.542\times {10}^{-62}}$ =

1.22 $\times {10}^{-15}$ s

For level $n_3$ = 3, $T_3$ = $\frac{2\times n_3^2\times h^2\times {?}_0}{v_3\times m\times e^2}$ = $\frac{2\times 9\times {(6.626\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{0.727\times {10}^6\times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$ = $\frac{6.993\times {10}^{-77}}{1.694\times {10}^{-62}}$ = 4.13 $\times {10}^{-15}$ s

12.6 For ground level, $n_1$ = 1

Let $E_1$ be the energy level at $n_1$ . From the relation

$E_1=\frac{E}{n_1^2}$ where E = -13.6 eV, we get

$E_1=\frac{E}{n_1^2}$ eV = -13.6 V

For higher level, $n_2$ = 4

Let $E_2$ be the energy level at $n_2$ . From the relation

$E_2=\frac{E}{n_2^2}$ where E = -13.6 eV, we get

$E_2=\frac{-13.6}{4^2}$ eV = -0.85 V

The amount of energy absorbed by proton is given as $E_p$ = $E_2-E_1$ = -0.85 + 13.6 eV = 12.75 eV = 12.75 $\times 1.6\times {10}^{-19}$ J = 2.04 $\times {10}^{-18}$ J

For a photon of wavelength $?,$ the expression of energy is written as

$E_p$ = $\frac{hc}{?}$ , where

c = speed of light = 3 $\times {10}^8$ m/s

h = Planck’s constant = 6.626 $\times {10}^{-34}$ Js

We get $?=\frac{hc}{E_p}$ = $\frac{6.626\times {10}^{-34}\times 3\times {10}^8\mathrm{}}{2.04\times {10}^{-18}}$ = 9.744 $\times {10}^{-8}$ m = 97.44 nm

The frequency of the proton is given by,

$?=\frac{c}{?}$ = $\frac{3\times {10}^8}{9.744\times {10}^{-8}}$ Hz= 3.1 $\times {10}^{15}$ Hz

12.16 We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck's constant (h).

The angular momentum of the Earth in its orbit is of the order of  ${10}^{70}$ h. This leads to a very high value of quantum levels n of the order of ${10}^{70}$ . For large values of n, successive energies and angular momentum are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

12.15 Total energy of the electron, E = -3.4 eV

Kinetic energy of the electron is equal to the negative of the total energy

K = -E = 3.4 eV

Potential energy (U) of the electron is equal to twice the negative of kinetic energy

U = -2K = -6.8 eV

The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change.

12.13 It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n-1)

We have the relation for energy ( $E_1$ ) of radiation at level n as:

$E_1$ = h ${?}_1$ = $\frac{hm{e}^2}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2}$ $\times \frac{1}{{\left(n\right)}^2}$ ………………(i)

Where,

${?}_1$ = Frequency of radiation at level n

h = Planck’s constant

m = mass of hydrogen atom

e = charge of an electron

${?}_0$ = Permittivity of free space

Now, the relation for energy ( $E_2$ ) of radiation at level (n-1) is given as:

$E_2$ = h ${?}_2$ = $\frac{hm{e}^2}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2}$ $\times \frac{1}{{(n-1)}^2}$ ………………(ii)

Where,

${?}_2$ = Frequency of radiation at level (n-1)

Energy (E) released as a result of de-excitation:

E = $E_2-E_1$

$h?$ = $E_2-E_1$ …………………………(iii)

where,

$?$ = Frequency of radiation emitted

Putting values of equation (i) and (ii) in equation (iii), we get,

$?$ = $\frac{m{e}^2}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2}$ $\left[\frac{1}{{(n-1)}^2}-\frac{1}{{\left(n\right)}^2}\right]$

= $\frac{m{e}^2(2n-1)}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2{n}^2{(n-1)}^2}$

For large n, we can write (2n-1) = 2n and (n-1) = n

Therefore, $?$ = $\frac{m{e}^2}{32{?}^3{?}_0^2({\frac{h}{2?})}^3{n}^3}$ ………..(iv)

Classical relation of frequency of revolution of an electron is given as:

${?}_c$ = $\frac{v}{2?r}$ …………………..(v)

Velocity of the electron in the $n^{th}$ orbit is given as:

v = $\frac{e^2}{4?{?}_0\left(\frac{h}{2?}\right)n}$ ……………...(vi)

And, radius of the $n^{th}$ orbit is given as:

r = $\frac{4?{?}_0({\frac{h}{2?})}^2}{m{e}^2}{n}^2$ ……………….(vii)

By putting the values of (vi) and (vii) in equation (v), we get

${?}_c$ = $\frac{m{e}^2}{32{?}^3{?}_0^2({\frac{h}{2?})}^3{n}^3}$

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

12.12 Radius of the first Bohr orbit is given by the relation:

$r_1$ = $\frac{4?{?}_0{\left(\frac{h}{2?}\right)}^2}{m_e{e}^2}$ ………………(1)

Where,

${?}_0=$ Permittivity of free space

h = Planck’s constant = 6.626 $\times {10}^{-34}$ Js

$m_e$ = mass of electron = 9.1 $\times {10}^{-31}$ kg

e = Charge of electron = 1.9 $\times {10}^{-19}$ C

$m_p$ = mass of a proton = 1.67 $\times {10}^{-27}$ kg

r = distance between the electron and proton

Coulomb attraction between an electron and a proton is given as:

$F_c$ = $\frac{e^2}{4?{?}_0{r}^2}$ ……………….(2)

Gravitational force of attraction between an electron and a proton is given as:

$F_G$ = $\frac{G{m}_p{m}_e}{r^2}$ ……………….(3)

Where, G = Gravitational constant = 6.67 $\times {10}^{-11}$ N $m^2$ / ${kg}^2$

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write

$F_c=F_G$

$\frac{e^2}{4?{?}_0{r}^2}=\frac{G{m}_p{m}_e}{r^2}$

or

$G{m}_p{m}_e=\frac{e^2}{4?{?}_0}$

By putting the above value in equation (1), we can write

$r_1$ = $\frac{4?{?}_0}{e^2}\times \frac{{\left(\frac{h}{2?}\right)}^2}{m_e}$ = $\frac{1}{G{m}_p{m}_e}$ $\times \frac{{\left(\frac{h}{2?}\right)}^2}{m_e}$ = $\frac{{\left(\frac{h}{2?}\right)}^2}{G{m}_p{m}_e^2}$

= $\frac{{\left(\frac{6.626\mathrm{}\times {10}^{-34}}{2?}\right)}^2}{6.67\mathrm{}\times {10}^{-11}\times 1.67\mathrm{}\times {10}^{-27}\times {9.1\mathrm{}\times {10}^{-31}}^2}$ = $\frac{1.112\times {10}^{-68}}{9.224\times {10}^{-98}}$ = 1.21 $\times {10}^{29}$ m

It is known that the universe is 156 billion light years wide or 1.5 $\times {10}^{27}$ m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

12.11 About the same - The average angle of deflection $?$ -particles by a thin gold foil predicted by Thomson's model is about the same size as predicted by Rutherford's model. This is because the average angle was taken in both models.

Much less - The probability of scattering of $?$ -particles at angles greater than 90 $°$ predicted by Thomson's model is much less than that predicted by Rutherford's model.

Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

Thomson's model - It is wrong to ignore multiple scattering in Thomson's model for the calculation of average angle of scattering of $?$ -particles by a thin foil. This is because a single-collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

12.14 Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Mass of an electron, $m_e$ = 9.1 $\times {10}^{-31}$ kg

Speed of light, c = 3 $\times {10}^8$ m/s

Let us take a quantity involving the given quantities as $\left(\frac{e^2}{4?{?}_0{m}_e{c}^2}\right)$

Where

${?}_0$ = permittivity of free space and

$\frac{1}{4?{?}_0}$ = 9.1 $\times {10}^9$ N $m^2{C}^{-1}$

Hence, $\left(\frac{e^2}{4?{?}_0{m}_e{c}^2}\right)$ = 9.1 $\times {10}^9\times \frac{{(1.6\mathrm{}\times {10}^{-19})}^2}{9.1\mathrm{}\times {10}^{-31}\times {(3\mathrm{}\times {10}^8)}^2}$ = 2.844 $\times {10}^{-15}$ m

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Mass of an electron, $m_e$ = 9.1 $\times {10}^{-31}$ kg

Planck’s constant, h = 6.623 $\times {10}^{-34}$ Js

Hence, the value of the quantity taken is in the order of the atomic size.

12.2 In the alpha-particle scattering experiment, if a thin sheet of hydrogen is used in place of a gold film, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 $*{10}^{-27}$ kg) is less than the mass of incident  $?-$ particles (6.64 $*{10}^{-27}$ kg). Thus, the mass of the scattering particles is more than the target nucleus (hydrogen). As a result, the $?-$ particles would not bounce back if solid hydrogen is used in the $?-$ particle scattering experiment.

12.17 Mass of a negatively charged muon, $m_{?}$ = 207 $m_e$

According to Bohr’s model

Bohr radius, $r_e$ $?$ $\frac{1}{m_e}$

And, energy of a ground state electronic hydrogen atom $E_e$ $?$ $m_e$

Also, the energy of a ground state muonic hydrogen atom, $E_u$ $?$ $m_u$

We have the value of the first Bohr orbit, $r_e$ = 0.53 Å = 0.53 $\times {10}^{-10}$ m

Let $r_o$ be the radius of muonic hydrogen atom

At equilibrium, we can write the relation as:

$m_e{r}_{?}$ = $m_e{r}_e$

207 $m_e{r}_{?}$ = $m_e{r}_e$

$r_{?}=\frac{r_e}{207}$ = $\frac{0.53\mathrm{}\times {10}^{-10}}{207}$ =2.56 $\times {10}^{-13}$ m

Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 $\times {10}^{-13}$ m

We have $E_e$ = -13.6 eV

12.3 Rydberg’s formula is given as:

$\frac{hc}{?}$ = 21.76 $\times {10}^{-19}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Where, h = Planck’s constant = 6.6 $\times {10}^{-34}$ Js

c = speed of light = 3 $\times {10}^8$ m/s

$?$ = Wavelength

$n_1$ and $n_2$ are integers.

The shortest wavelength present in the Paschen series of the spectral lines is given for the values $n_1=3$ and $n_2=?$

Therefore, $\frac{hc}{?}$ = 21.76 $\times {10}^{-19}\left[\frac{1}{3^2}-\frac{1}{{?}^2}\right]$

$\frac{hc}{?}=$ 2.42 $\times {10}^{-19}$

$?=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.42\times {10}^{-19}}$ = 8.189 $\times {10}^{-7}$ m= 818.9 nm